An alpine rescue team is using a slingshot to send an emergency medical packet to climbers stranded on a ledge. (height=252 m, horizontal distance=416 m, landing angle= 75° below the horizontal).

Question

What should be the launch speed from the slingshot?
What is the launch angle?
What is the time of flight?
What is the highest point reached by the packet?

Answer 1

you know hf, distance final, df, and the ratio of Vverticalfinal to vhoriz final frm the landing angle. This is going to be an exercise in math.

Lets work it backwards. Assume V as the launching velocity on the ledge, as if they sent a package to you.

hf=hi+ViSin75*time-4.9t^2 or
-252=V Sin 75*t-4.9t^2

in the horizontal
416=Vcos75*t solve for t
t= 416/Vcos75 now put that into the first equation for t, and then solve for V. A bit of algebra is required, get a pad of paper.

Now having V at launch, you have V at the landing (from initial KE+initial PE= final KE)

YOu have the final V, and the horizontal V, so you can find the final angle (theta=arcCos Vhoria/V)

you should be able to take it from here.
for highest point, you have the initial KE, the initial PE, that should equal the KE fromhorizonal+ pE at the top. find max height from that.